Teddy Einstein Math 4320


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1 Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective homomorphism G G because compositions of homomorphisms are homomorphisms and compositions of bijections are bijections. Hence f g Aut G. If id G G G is defined by g g, it is trivially an automorphism. For f Aut G, f id G = f = id G f. Thus the identity map on G is the identity in Aut G. Recall that if G, H are groups and q G H is an isomorphism, then its inverse function f 1 H G is an isomorphism as well. Thus if f G G is an automorphism, its inverse function f 1 G G is also an automorphism. Hence Aut G is closed under inverses. Function composition is associative, so we have associativity. Thus Aut G satisfies the axioms of a group. ii. Prove that γ G Aut G defined by g γ g where γ g h g 1 hg is a homomorphism. Proof. Let g, k G. Observe that for h G: γ(gh)(k) = (gh) 1 k(gh) = h 1 g 1 kgh = h 1 (g 1 kg)h = γ h (γ g (k)) = [γ(h) γ(k)](h) so γ h is a homomorphism of groups. iii. Prove that ker γ = Z(G). Proof. Observe that ker γ = {g G γ g = id G } = {g G g 1 xg = x x G} = {g G xg = gx x G} = Z(G), yielding the desired result. iv. Prove Im γ Aut G. Proof. Let g G so that γ g is an arbitrary element of Im γ. We need to prove that given f Aut G, f 1 γ g f Im γ. Now we see that for k G: f 1 γ g f(k) = f 1 (g 1 f(k)g) = f 1 (g 1 ) f 1 f(k) f 1 (g) = (f 1 (g)) 1 ck 9 f 1 (g) = γ f 1 (g)(k) using the fact that f 1 is a homomorphism. Hence f 1 γ g f = γ f 1 (g) Im γ, so Im γ Aut G. 1
2 2 Problem 2: (20 pts) Let G be a group with x, y G. Define the commutator [x, y] = x 1 y 1 xy and the commutator subgroup of G = x 1 y 1 xy x, y G G. 1 i. Prove that G G. Proof. Let x, y G and h G. Observe that: h 1 x 1 y 1 xyh = h 1 x 1 y 1 hh 1 xyh = (h 1 x 1 hh 1 y 1 h)(h 1 xhh 1 yh) = [h 1 xh, h 1 yh] because for any k G, (h 1 kh) 1 = h 1 k 1 h. Hence the conjugate of any commutator is contained in G. Now let α G be a product of a commutators α i G so that: α = α 1 α 2... α n Observe that for g G, g 1 α i g G by the preceding. Now we see that: g 1 αg = g 1 α 1 α 2... α n g = g 1 α 1 g 1 gα 2 g 1 gα 3 g 1 g... gg 1 α n g = n i=1 g 1 α i g where each g 1 α i g G, so g 1 αg is a product of commutators. Thus g 1 αg G. Hence G G. ii. Prove that G/G is abelian. Proof. Let x, y G. Observe that the coset x 1 y 1 xyg = G because x 1 y 1 xy G. Hence xg yg = xyg = yxg = yg xg, so G/G is abelian. iii. If ϕ G A is a homomorphism, where A is an abelian group, prove that G ker ϕ. Conversely, if G ker ϕ, prove that Im ϕ is abelian. Proof. Suppose x, y G and let ϕ G A be a homomorphism where A is abelian. Observe that ϕ(x 1 y 1 xy) = ϕ(x) 1 ϕ(y) 1 ϕ(x)ϕ(y). Since A is abelian, we can rearrange to find: ϕ(x 1 y 1 xy) = ϕ(x) 1 ϕ(x)ϕ(y) 1 ϕ(y) = 1 A so x 1 y 1 xy ker ϕ. Thus every commutator lies in ker ϕ which is a subgroup of G, so G, the smallest subgroup of G containing all commutators of G, must be a subgroup of ker ϕ. Now conversely, suppose G ker ϕ where ϕ G H is a group homomorphism. By the first isomorphism theorem, Im ϕ G/ ker ϕ. Thus it suffices to show G/ ker ϕ is abelian. For x, y G: x 1 y 1 xy ker ϕ = ker ϕ 1 Note that the commutator subgroup contains ALL (finite) PRODUCTS of commutators which may not be commutators themselves.
3 3 because x 1 y 1 xy ker ϕ since G ker ϕ. Hence we have by properties of quotient groups: xy ker ϕ = yx ker ϕ x ker ϕy ker ϕ = y ker ϕx ker ϕ which shows that G/ ker ϕ is abelian. Hence Im ϕ is abelian. iv. If G H G, prove that H G. Proof. By proposition 2.123, H G if and only if H/G is a normal subgroup of G/G. Observe that G/G is abelian and every subgroup of an abelian group is normal. 2 Thus H/G is normal in G/G which is abelian by the previous part. We then conclude that H is normal in G. Problem 3: Supplemental Let (G, ) and (H, ) be groups and let f H Aut G be a homomorphism. operation on the set G H by: Define a binary (g 1, h 1 ) (g 2, h 2 ) = (g 1 f(h 1 )(g 2 ), h 1 h 2 ) Prove that (G H, ) is a group. Proof. Let g, g G and h, h H. We see that f(h)(g ) G because f(h) Aut G. Thus by closure of G, we have g f(h)(g ) G. Similarly, h 1 h 2 H, so: (g, h) (g, h ) = (g f(h)(g ), h 1 h 2 ) G H so (G H) is closed under. Observe that: (1 G, 1 H )(g, h) = (1 G f(1 H )(g), 1 H h) = (1 G g, 1 H h) = (g, h) (g, h)(1 G, 1 H ) = (g f(h)(1 G ), h 1 H ) = (g 1 G, h 1 H ) = (g, h) making (1 G, 1 H ) the identity in (G H, ). Note that f(1 H ) = id G because f is a homomorphism (which means identity maps to identity), and f(h)(1 G ) = 1 G because f(h) is a homomorphism which means it takes identity to identity as well. For inverses, if we want (g, h) (a, b) = (1 G, 1 H ), we need (g f(h)a, h b) = (1 G, 1 H ). Thus we need b = h 1 and f(h)(a) = g 1. Of course, such an a G exists because f is an automorphism and is surjective. Thus setting a, b with these parameters, we obtain an inverse for (g, h). Finally, we need to check associativity of the operation. Let g G and h H. Then check: (g, h) ((g, h ) (g, h )) = ((g, h) (g, h )) (g, h ) this is annoying but straightforward and is hence omitted. 2 This is easy, if A is an abelian group, for all a, b A, a 1 ba = b, so every subgroup is fixed under conjugation.
4 4 Problem 4: Supplemental Denote the group constructed in the previous exercise by G f H. For all n 3 find f n such that D 2n I n fn I 2. Proof. Recall that D 2n = a, b a n = b 2 = 1, ab = ba 1. With this presentation, every element of D 2n can be written in the form a k b m where 0 k < n, 0 b < 2. 3 Naively, we would want (k, m) I n I 2 to correspond to a k, b m where denotes the equivalence class of the specified integer either mod 2 or mod n as appropriate. When multiplying together a k b m a k b m, if m = 0, then we get a k+k b m+m. Otherwise if m = 1, we get a k a k bb m = a k k b m+m. In other words, if we want to realize D 2n in the desired form, we will want to apply some kind of inversion when the second coordinate of the first term is 1. Specifically, suppose we have an f I 2 Aut I n giving the desired transformation and let x, y, z, w Z. We want (x, y)(z, w) = (x+f(y)(z), y+w). In order for the corespondence we want, we need x+f(y)(z) = x + z if y = 0 and x z if y = 1. Hence define f Z Aut I n by: 4 f (t) = h ( 1) t h thus f (t) is either the identity or inversion, so f (t) is always an automorphism of I n. Check f (t + t ) = (h ( 1) t+t h) = (h ( 1) t ( 1) t h) = (f (t) f (t )). Hence f is a homomorphism whose kernel is precisely 2Z, then even numbers. Since I 2 = Z/2Z, f (t) induces a well defined homomorphism f I 2 Aut I n such that f(0) is identity and f(1) is inversion. Now set our group G = I n f I 2. Define a map ϕ D 2n G defined by ϕ a k b m (k, m). We need to check that ϕ is a homomorphism. Let k, m, k, m be integers with 0 k, k < n and m, m {0, 1}. Suppose first that m is even: ϕ(a k b m )ϕ(a k b m ) = (k, 0)(k, m ) = (k + f(0)(k ), 0 + m ) = (k + k, m + m ) On the other hand if m is odd: = (k + k, m + m ) = ϕ(a k+k b m+m ) ϕ(a k b m )ϕ(a k b m ) = (k, 1)(k, m ) = (k + f(1)(k ), 1 + m ) = (k k, 1 + m ) = ϕ(a k k b 1+m ) = ϕ(a k b 1 a k b m ) = ϕ(a k b m a k b m ) Thus ϕ is a homomorphism. If ϕ(a m b k ) = (0, 0), then with 0 k < n, 0 m < 2, then m = 0, k = 0, so a m b k = e, so ϕ is injective. Trivially, (m, k) = ϕ(a m b k ), so ϕ is surjective. Thus ϕ is an isomorphism of groups. 3 In order to see this, observe that whenever a b appears to the left of a power of a, the relation ab = ba 1 allows it to be moved to the right of any power of a. 4 Note that we are writing In as an additive group here with operation +.
5 5 Problem 5: Prove that the translation τ a S G defined by τ a a ag is a regular permutation (Assume G is finite). Proof. We need to show that τ a is a product of disjoint cycles of the same length with no fixed points or is the identity when S G is identified with S G. Observe that the elements can be reached from g by applying τ a repeatedly are exactly those of the form (τ a ) n (g) = a n g for n N where in practice, a a = 1, so this consists exactly of the elements g, ag, a 2 g,..., a a 1 g. Hence in the disjoint cycle decomposition of τ a, g is in the cycle (a, a 2 g, a 3 g,..., a a 1 g). Note that this cycle coincides exactly with a g, a right coset of g in G. Since a was arbitrary, any element in G appears in the disjoint cycle decomposition of τ a in a cycle which is some permutation of the elements of a g. Hence the cycle length for g is exactly a g. The proof to Lagrange s theorem shows that all cosets of a have the same order, namely a = a. Thus we see that τ a is a product of disjoint cycles of equal length. If τ a has a fixed point, then ag = g for some g G, but then a = 1 G. In that case, τ a is the identity map on G, so τ a has a fixed point only if it is the identity. Hence τ a is a regular permutation. Problem 6: Let σ, τ S 5 with σ a 5 cycle and τ a transposition. Show that σ, τ = S 5. Proof. Without loss of generality, we may relabel so that τ = (12). Since σ is a 5cycle, then there exists 0 < k < 5 such that σ k (1) = 2. Since 5 is prime and 0 < k < 5 so that (5, k) = 1, σ k is also a 5 cycle. Thus (σ k ) l (2) 1, 2 if l < 3. Since we only needed to fix the labelling of 1, 2, we can freely change the labelling of 3, 4, 5 and still have a group isomorphic to S 5 with τ = (12). Thus we can relabel so that without loss of generality σ k = (12345) because the first two letters of the 5 cycle σ k are fixed as 1, 2, so the last 3 must be among 3, 4, 5. Now observe that the element (12345) n (12)(12345) n = (1 + n, 2 + n) for 0 n 3 and is in σ, τ (You should actually show this calculation). Thus σ, τ contains all of the simple transpositions in S 5. In a previous homework exercise, we showed that the simple transpositions in S n generate S n, so σ, τ must contain S 5, so we conclude they are equal.
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